Let us do k=2 and you try k=3. First pick 2 digits numbers that you are going to use. There are ways of doing that. But once you have choose these two numbers there are ways of ordering them. Multiply those two together and you get your answer.

I have the exact same question, but I have to do it for all 6 values of k. I understand why there are ways to pick 2 digits, but why are there ways to order them? I've been trying to work the problem out this way:
For each value of k, treat the number of possible numbers as a 6-digit base-k number, subtract the numbers that don't have k distinct digits, and finally multiply it by the number of possible values for each distinct digit. So, for example, if k=4, I'd work it out as:
The number of numbers in a base 4, 6-digit number, minus all the numbers with only 3 distinct digits, 2 distinct digits, and 1 distinct digit, and multiply it by the number of ways to choose those 4 digits.
I know my reasoning is wrong, because when I try the same idea for k=5, I get a value over 1 million, which is obviously impossible.

So how do you calculate the number of ways to order the digits? I get that 6! is because it's a 6-digit number, but what the denominator?